Let $f(n)$ be the base-10 logarithm of the sum of the elements of the $n$th row in Pascal's triangle. Express $\frac{f(n)}{\log_{10} 2}$ in terms of $n$. Recall that Pascal's triangle begins

\begin{tabular}{rccccccccc}
$n=0$:& & & & & 1\\\noalign{\smallskip\smallskip}
$n=1$:& & & & 1 & & 1\\\noalign{\smallskip\smallskip}
$n=2$:& & & 1 & & 2 & & 1\\\noalign{\smallskip\smallskip}
$n=3$:& & 1 & & 3 & & 3 & & 1\\\noalign{\smallskip\smallskip}
$n=4$:& 1 & & 4 & & 6 & & 4 & & 1\\\noalign{\smallskip\smallskip}
& & & & & $\vdots$ & & & &

\end{tabular}
Solution: Computing the sums of the entries in the first few rows suggestions that the sum of the entries in row $n$ is $2^n$. Indeed, one way to prove this formula is to note that the $k$th entry of the $n$th row is $\binom{n}{k}$ (if we say that the entries in the $n$th row are numbered $k=0,1,\dots,n$). We have  \[
\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\dots +\binom{n}{n} = 2^n,
\]since both sides calculate the number of ways to choose some subset of $n$ objects. It follows that $f(n)=\log_{10} (2^n)$, which means that $\frac{f(n)}{\log_{10} 2}=\frac{\log_{10} (2^n)}{\log_{10} 2}$. Applying the change of base formula gives us $\log_2 (2^n)=\boxed{n}$.